Determine which condition of Rolle's theorem fails

Determine which condition of Rolle's theorem fails for $f(x) = \bigl|\sin x\bigr|$ on $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$.

Step 1: Continuity holds

$\sin x$ is continuous and so is the absolute value, so their composition is continuous everywhere.

Step 2: The endpoint values agree

\[ f\left(-\frac{\pi}{2}\right) = \bigl|-1\bigr| = 1, \qquad f\left(\frac{\pi}{2}\right) = \bigl|1\bigr| = 1. \]

Step 3: Where differentiability fails

On the interval $\sin x$ is negative to the left of $0$ and positive to the right, so

\[ f(x) = \begin{cases} -\sin x, & -\tfrac{\pi}{2}\lt x\lt 0,\\ \ \ \sin x, & 0\lt x\lt \tfrac{\pi}{2}, \end{cases} \qquad f'(x) = \begin{cases} -\cos x, & x\lt 0,\\ \ \ \cos x, & x>0. \end{cases} \]

At the origin the difference quotient has different one-sided limits: $\lim_{h\to0^{+}}\frac{\sin h}{h} = +1, \qquad \lim_{h\to0^{-}}\frac{-\sin h}{h} = -1$.

Since $+1\neq-1$, the derivative $f'(0)$ does not exist.

Step 4: The conclusion fails too

For $x\neq0$ in the interval, $f'(x) = \pm\cos x$, and $\cos x\neq0$ there; at $x = 0$ the derivative does not exist. So no $\xi$ satisfies $f'(\xi) = 0$.

Step 5: Verification

\[f'\left(-\frac{\pi}{4}\right) = -\frac{\sqrt2}{2} \approx -0.707, \qquad f'\left(\frac{\pi}{4}\right) = +\frac{\sqrt2}{2} \;\;\checkmark\]

\[\text{Neither vanishes anywhere on the interval, and } \bigl|f'\bigr|\ge\cos\frac{\pi}{2} = 0 \text{ only in the limit} \;\;\checkmark\]

\[\text{Minimum of } f \text{ is } 0 \text{ at } x = 0, \text{ attained at the non-differentiable point} \;\;\checkmark\]